Question

If ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left({\log }_{e}a+{\log }_{e}b\right)$ then relation between a and b will be

Answer 1

We have,

${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left({\log }_{e}a+{\log }_{e}b\right)$

${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left({\log }_{e}ab\right)$ $\because \log m+\log n=\log mn$

${\log }_e\left(\frac{a+b}{2}\right)=\left({\log }_e\sqrt{ab}\right)$ $\because n\log m=\log m^n$

$\frac{a+b}{2}=\sqrt{ab}$

$a+b=2\sqrt{ab}$

$a+b-2\sqrt{ab}=0$

$(\sqrt{a}-\sqrt{b}{)}^2=0$

$\sqrt{a}-\sqrt{b}=0$

$\sqrt{a}=\sqrt{b}$

Hence, this is the answer.